For any given integer B
greater than 1, each natural number N has a unique expression of the form

where each of the digits d0,
d1,... is an integer in the range from 0 to B-1. It's
conventional to omit the operation symbols and powers of B, and simply
express the number N by its sequence of digits, arranged in descending
order. Thus the number N would be expressed as the string of digits ...d2d1d0.
Typically the left-most digit is the largest non-zero digit. This form of
expression is called place-value notation, and the integer B is called the
base (or radix). Obviously we have d0 = N modulo B, so if we are
working in the modulus B we can simply replace each number by its least
significant digit.

The base-B digits of a
number also satisfy some interesting relations modulo B-1. If we let SB(N)
denote the sum of the base-B digits of N, we have

Since each of the
quantities Bk-1 is divisible by B-1, we see that N is congruent to SB(N) modulo B-1. (This is also obvious from the fact that B = 1 modulo B-1, and so
every power of B is also congruent to 1, and thus the base-B representation
of a number modulo B-1 reduces to just the sum of its digits.) It follows
that N is also congruent to SB(SB(N)), and so on. Thus
to find the residue class of N modulo B-1 we need only sum the base-B digits
of N, and then sum the base-B digits of the resulting number, and so on,
until reaching a single digit result. Equivalently, we can just evaluate
modulo B-1 the sum of the digits of N. This shows that if xy = z for two
integers x,y, the congruence xy = z (mod B-1) can be re-written in terms of
the residues modulo B-1 as

We can verify this
explicitly by simply multiplying out the base-B representations

The base-B digits of z = xy
are then related to the coefficients of this product by

and so on, where c0,
c1,... are the "carry" terms in the multiplication.
(The carry terms need not be in the range from 0 to B-1.) Adding all these
expressions together, we get

This confirms the previous
congruence modulo B-1, and also gives a simple formula for the sum of the
carries in a multiplication in terms of the sums of digits of the operands
and product.

This is the basis for the
old technique of "casting out nines", which was sometimes useful
for checking multiplications performed by hand. To illustrate, suppose we
multiply 23579 by 84427 to give the product 199074233. Needless to say, it's
easy to check our calculation modulo 10 by noting that the residues of the
operands are 9 and 7 (mod 10), giving a product of 63, whose residue is 3
(mod 10), consistent with the fact that the residue of 199074233 (mod 10) is
3. This check just makes use of the least significant digit of each number,
so it is not very robust.

For a stronger test, we can
verify our calculation modulo 9, which can be done easily by means of the congruences
modulo B-1 described above. We first find the residue class of 23579 (mod
9), which we can do by simply adding up the digits and discarding all
multiples of 9. This is why the method is called "casting out
nines". The number 23579 has the digits 2, 3, 5, 7, and 9. We can
obviously ignore the digit 9, and we can discard the digits 2 and 7 because
they sum to 9. This leaves just 3 + 5 = 8, so we know that 23579 is
congruent to 8 (mod 9). Likewise we can sum the digits of 84427, casting out
all whole multiple of 9 along the way, to give the result 7. Now, the
product of 8 and 7 is 56, and the sum of those digits is 5+6=11, and the sum
of those digits is 1+1=2. To check the correctness of our multiplication, we
just need to verify that the sum of the digits of the product 199074233,
after casting out all multiples of 9, is 2.

The combination of the
checks modulo B and modulo B-1 assures us that the product is correct modulo B(B-1),
which is 90 in decimal arithmetic. The chance of a random error passing both
of these tests is about 1/90. With roughly the same amount of labor we can
use a process of “casting out elevens” to perform a similar test modulo B+1
by summing the base-B digits of the number with alternating signs. For
example, the decimal number 73875613 is congruent modulo 11 to 3 – 1 + 6 – 5
+ 7 – 8 + 3 – 7. To perform this summation in our heads, we can sum the
positive terms 3+6+7+3, casting out elevens as we go, giving the result 8.
(We could also just sum these numbers to give 19, and then repeat to give 9 –
1 = 8.) Then we can sum the negative terms 1+5+8+7, casting out elevens, to
give 10. The overall residue modulo 11 is therefore 8 – 10, which is -2,
congruent to 9 mod 11.

As an illustration of the
occasional usefulness of "casting out nines", consider social
security numbers, which can be regarded as consisting of a three-digit
integer, a two-digit integer, and a four-digit integer. For example, one
possible social security number is 728;35;4589. (We use semi-colons to avoid
confusion between the usual dashes and minus symbols.) Do there exist any
social security numbers such that each of the numerals 1 through 9 appears
exactly once, and such that the four-digit number is the product of the
three-digit and two-digit numbers?

We can use modulo 9
arithmetic to simplify the search for solutions. It's useful to recall the
multiplication table modulo 9.

Let's denote a social
security number by X;Y;Z, where X is a three-digit integer, Y is a two-digit
integer, and Z is a four-digit integer. Since we require that each of the
numerals 1 through 9 appears exactly once, we have X + Y + Z = 0 (mod 9). We
also require XY = Z (mod 9). Eliminating Y from these two congruences, we
have

It follows that, for any
given residue Z (mod 9), we have

This has integer solutions
if and only if Z2 - 4Z is a square (mod 9). The
only squares modulo 9 are 0, 1, 4, and 7, so the only possible residue
classes for Z are 0 and 4. Noting that 0 has three square roots (0,3,6)
modulo 9, each of these values of Z gives three sets of X and Y values, as
listed below.

Now we need only consider
each of these six possibilities in turn to determine all possible solutions.
Taking the case X = Y = Z = 0 (mod 9), we know Y must be one of the numbers
18, 81, 27, 72, 36, 63, 45, or 54. If Y = 18, we know X and Z must be
composed of the digits 2, 3, 4, 5, 6, 7, and 9, each appearing exactly once,
such that X = Z = 0 (mod 9). Hence X must be composed of the digits in one
of the sets {2,7,9}, {3,6,9}, {4,5,9}, {5,6,7}. Focusing on the first set,
{2,7,9}, it's clear that the first digit must be 2 (because X cannot exceed
700 without Z exceeding 9999), and the last digit cannot be 9 (because the
product of numbers ending with 8 and 9 ends in 2, which means the numeral 2
would be repeated). Therefore, the only possibility for the set {2,7,9} is X
= 297. If we then multiply this out, we get (297)(18) = 5346, which is
indeed a solution. We can quickly rule out the remaining three possible sets
of numerals for X, so the only solution with X = Y = Z = 0 (mod 9) and Y = 18
is the social security number 297;18;5346. Clearly we can rule out Y = 81
based on size alone, so the next possible case is Y = 27, which means X must
be composed of the digits in one of the sets {1,8,9}, {3,6,9}, {4,5,9},
{4,6,8}. Focusing on {1,8,9}, the first digit must be 1, so the only
possibilities are 198 and 189. Indeed we find that (198)(27) = 5346, so this
is another solution. The other three digit sets can be immediately ruled out
for X due to size. Continuing on in this way, we can show that none of the
remaining values of Y leads to a complete solution, so there are just two
solutions of the form X = Y = Z = 0 (mod 9).

Applying the same sort of
analysis of the case X=6, Y=3, Z=0 (mod 9), we find that there are exactly
three solutions of this form, namely 159;48;7632, 186;39;7254, and 483;12;5796.
The case X=3, Y=6, Z=0 (mod 9) yields just one solution, namely 138;42;5796.
With Z=4 (mod 9), only the case X=4,Y=1, Z=4 (mod 9) yields a solution,
namely 157;28;4396. Thus we have determine that there are exactly seven
social security numbers with the specified properties:

It's interesting that this
set contains two pairs with the same last four digits.